# Op-Amp question

Tags:
1. Dec 27, 2014

### GBA13

1. The problem statement, all variables and given/known data
Hi Guys,

I am trying to solve this question, please look at the attached picture

2. Relevant equations
The general equation for a inverting amp is -Rf/R1 * Vin = Vout

3. The attempt at a solution
Well as the question says the two resistors, R2 and R must be treated as parallel so I think the best way to do it would be to use the three resistors, R1, R2, R to find an equivalent resistance for R1. I thought I could do this by thinking that R1 is in series with the parallel combination on R2 and R but when I do the calculation and use the general equation I get the wrong answer which is 20k Ohms . Can anyone offer a hand?

Thanks!

#### Attached Files:

• ###### Screen Shot 2014-12-27 at 19.40.54.png
File size:
23 KB
Views:
119
2. Dec 27, 2014

### Staff: Mentor

You cannot treat them as parallel resistors everywhere because the current going through the 50kOhm resistor goes through R2 but not through R.
I would start at that point: you can calculate the current through R2 and the corresponding voltage as function of the output voltage. Then use Kirchhoff's law at the node between R/R1/R2 and the "design" input voltage to solve for R.

Edit: Hmm, I did that and got 20 kOhm... do you know the answer that is supposed to be right?

3. Dec 27, 2014

### GBA13

Hi mfb, thanks for answering. Just to clarify the correct answer is 20kOhm, I phrased that part of the question badly!
Ok that makes sense. As no current can flow into the input isn't the current going across the 50kOhm resistor same as the current across R2?

I'm not sure how to find the current through R2. I would think that the current through R1 is Vin/10k so then the current through R2 would be (Vin/10k) - (Vin/R)? I'm not really sure if thats right but I'm not really sure how it would help me either...

4. Dec 27, 2014

### Staff: Mentor

Ah okay.

Right.

Why?
Even if the first thing would be right this would be wrong, and I don't see where you got that from.

5. Dec 27, 2014

### GBA13

Ok I guess that's just completely wrong.

To be honest I'm really not sure what to do, the voltage at point B is 0V so that makes me think that the voltage drop across the 50kOhm resistor must be 0V but then the current through it would be 0A so then the current in R2 would also be 0A so that can't be right.

6. Dec 27, 2014

### Staff: Mentor

If you have 0 V at the left side and Vout at the right side, what is the voltage drop across the 50kOhm resistor?

7. Dec 27, 2014

### Staff: Mentor

I am very much liking this angle of approach to the problem. Just thought I'd mention that :)

8. Dec 28, 2014

### GBA13

I would think you would have a voltage drop of Vout, which is -2Vin for the 50kOhm resistor.

9. Dec 28, 2014

### Staff: Mentor

Right. So what then is an expression for the current through the 50 k resistor? What other component does it pass through?

10. Dec 28, 2014

### GBA13

Ok so the current through resistor is -2Vin/50k and this current also goes through R2, so the current through R2 also has a current of -2Vin/50k. I think there would be a different current though R and R1 as there is a branch.

11. Dec 28, 2014

### GBA13

Thinking about it then the voltage in R2 would be (-2Vin/50k * 10k) so this must also be the voltage at R as they are in parallel. So im thinking the next thing would be to find the current in R..

12. Dec 28, 2014

### Staff: Mentor

Okay, so you have the current through R2 and the voltage at the top of R. You also know the input voltage, hence the voltage at both ends of R1..... What's the current through R1?

13. Dec 29, 2014

### GBA13

So the current would be ((-2Vin/50k * 10k) - Vin) /10k . I think I can then use KCL to find the current in the R branch.. so (-2Vin/50k * 10k) - Vin) /10k - (-2Vin/50k) is the current in R?

14. Dec 29, 2014

### Staff: Mentor

Yes, use KCL to find the current through R. Be careful about the sign of the voltage at the top of R. What direction is the current flowing through R2?